AP Chemistry Bonding Overview

I'm going to go ahead and draw a red line on the graph at 220 pometers. So pay close attention to the scale on the x-axis. So 220 pometers would be right here. And then it says that the average bond energy for this bond is 425 kJ per mole. So somewhere between 400 and 500 I would say that 425 kJ per mole would be located right here on the y ais. Now what I'm going to do is put a green dot on the graph at the intersection of those two red lines. [Music] Welcome to AP Chemistry Review and Practice. My name is Michael Farbaugh and I'll be your guide on the side as you strive toward that five on the AP Chemistry exam. Each episode will include answers and explanations to practice problems along with advice to help you avoid common mistakes. So, are you ready to review and practice for the AP Chemistry exam? I hope so, because this lesson starts right now. [Music] Welcome back to AP Chemistry Review and Practice. In this lesson, I'll give you the chance to review and practice topics 2.1 and 2.2. As you can see on this slide, these topics fall under unit 2 in the AP Chemistry curriculum, compound structure and properties. You can add your comments and questions below this video, and you can also email me with your comments and questions at a chemistry review and practice atgmail.com. Check out the links in the video description area, which include the link to the packet that accompanies this video. Now let's move on to the review for topic 2.1 types of chemical bonds. Here are the essential knowledge statements from the AP chemistry course and exam description. Electro negativity values for the representative elements increase going from left to right across a period and decrease going down a group. These trends can be understood qualitatively through the electronic structure of the atoms, the shell model and Kulum's law. Veence electrons shared between atoms of similar electro negativity constitute a non-polar covealent bond. For example, bonds between carbon and hydrogen are effectively non-polar even though carbon is slightly more electrogative than hydrogen. Veence electrons shared between atoms of unequal electro negativity constitute a polar covealent bond. The atom with a higher electro negativity will develop a partial negative charge relative to the other atom in the bond. In single bonds, greater differences in electro negativity lead to greater bond dipoles. All polar bonds have some ionic character and the difference between ionic and covealent bonding is not distinct but rather a continuum. The difference in electro negativity is not the only factor in determining if a bond should be designated as ionic or covealent. Generally bonds between a metal and non-metal are ionic and bonds between two non-metals are covealent. Examination of properties of a compound is the best way to characterize the type of bonding in a metallic solid. The veence electrons from the metal atoms are considered to be deoized and not associated with any individual atom. Now let's move on to the practice for topic 2.1. This is question one. It says a student assigned partial charges to each atom in each of the bonds shown below. Based on the electro negativity values of the atoms involved in each bond, circle all of the bonds for which the partial charges were labeled correctly by the student. So what we need to do to answer this question is consider that the atom with the higher electro negativity will develop a partial negative charge relative to the other atom in the bond. So these are the electro negativity values for various elements and you can see that the general trend is that electro negativity values tend to increase from left to right and they decrease from top to bottom. Now even though I'm showing you the actual electro negativity values on this slide, remember that you don't need to memorize these values for the AP chemistry exam. All you need to do is consider the general trends in electro negativity when comparing two different atoms. So the first two atoms that we're going to compare are arsenic as s and oxygen O. So we can see that oxygen is further to the top right corner of the periodic table compared to arsenic. So just based on the general trend, oxygen has a higher electro negativity than arsenic has. That means that the oxygen atom should have a partial negative charge. Now we'll see if the student assigned the partial charges correctly. Now they're putting a positive charge on oxygen instead of a negative charge. So that is incorrect. And now let's take a look at silicon and chlorine. Now we can see that silicon and chlorine are located in the same period. But since chlorine is further to the right, chlorine would have a higher electro negativity than silicon. So the chlorine atom should have a partial negative charge. Let's see if the student did this particular charge assignment correctly. And yes, there is a partial negative charge on the chlorine atom. Now let's move on to phosphorus and nitrogen. We can see that they are located in the same group. But since nitrogen is further to the top of the group, nitrogen should have a higher electro negativity than phosphorus has. So the nitrogen atom should have a partial negative charge and we can see that the student did also assign those charges correctly. Now moving on to a comparison of phosphorus and florine. So florine being further to the top right has a higher electro negativity than phosphorus has. So the florine atom should have a partial negative charge which the student also assigned correctly. Our next comparison is bromine and selenium. Now bromine is further to the right in that period compared to selenium. So bromine has a higher electro negativity than selenium. The bromine atom should have a partial negative charge which the student did not assign correctly. And then finally for chlorine and iodine which are located in the same group. Chlorine being further to the top of that group has a higher electro negativity than iodine. So the chlorine atom should have a partial negative charge which the student did not assign correctly. So these three circled bonds are the correct answers for question one in which all of the partial charges were labeled correctly. The atom with the higher electro negativity carries that partial negative charge in a polar covealent bond. Now let's take a look at question two. In question two, we are given three different chemical formulas and the question says which of the following correctly ranks the substances listed above in order of increasing ionic character in their bonding. So each compound contains the element chlorine CL, but we have silicon SI, magnesium Mg, and sulfur S. Now take a look on this slide. You can see that I've highlighted magnesium, silicon, sulfur, and chlorine which are all located in the same period of the periodic table. And you can also see that I've labeled these elements as either a metal, a non-metal or a metaloid. Now there is an important difference between a covealent bond and an ionic bond. And the usual method used in chemistry to classify a bond as either ionic or covealent is the following. metal and non-metal would lead to an ionic bond and two non-metals would lead to a covealent bond. Now this is a simple criterion that works well for many examples. However, the nature of a chemical bond is more complicated than this. In reality, all polar bonds have some ionic character and there is a continuum between the extremes of covealent bonding and ionic bonding. Now take a look on this slide and you'll see a link in the video description area to the FET simulation on molecule polarity. So what I'm doing is I'm moving that slider at the bottom and you can see that atom B which is labeled in green I'm increasing the electro negativity of that atom. So as I increase the electro negativity of atom B, I'm also increasing the magnitude of the bond dipole which is represented by that black arrow. So as the difference in electro negativity increases between the atoms, the bond character changes from more covealent to more ionic. So the ionic character of a bond increases with increasing differences between the electro negativities of the bonded elements. And that's how we're going to answer question two. So again all polar bonds have some ionic character and the difference between ionic and coalent bonding is not distinct but rather a continuum. Let's consider that the difference in electro negativity between magnesium and chlorine. You can see how far apart they are in the periodic table is greater than the difference in electro negativity between silicon and chlorine. Again, the difference in electro negativity between silicon and chlorine is greater than that between sulfur and chlorine. So, the sc2 compound has the lowest amount of ionic character and the mg2 compound has the greatest amount of ionic character. And it turns out that magnesium and chlorine are a metal and a non-metal. So magnesium and chlorine is more likely to form an ionic bond. So again, the correct answer is choice B. The ionic character of a bond increases with increasing differences between the electro negativities of the bonded elements. Now let's take a look at question three. Question three says, which of the following bonds are arranged in order of decreasing polarity? So this time we're looking for a smaller difference in electro negativity. In single bonds, greater differences in electro negativity will lead to greater bond dipoles. The greater the difference in electro negativity, the more polar the bond is. But in question three, we're looking for decreasing polarity. So we would like to see a decreasing difference in the electro negativities of the bonded atoms. Let's start with choice A. We have oxygen, nitrogen, and boron. And they are all bonded to the element florine. And you can see on the periodic table that oxygen and florine are closer together. And as I move from oxygen to nitrogen, now those two atoms are a little farther apart. And as I move from nitrogen to boron, those atoms are even farther apart, which means that would lead to a difference in electro negativity that's increasing. So in choice A, those bonds are actually becoming more polar, not less polar. So we can eliminate choice A. Now moving on to choice B, we have arsenic as bonded to either phosphorus, nitrogen or oxygen. So as I move from phosphorus to nitrogen to oxygen, I'm increasing the electro negativity of the atom that is bonded to arsenic. So again just like in choice A, as I move from arsenic phosphorus, arsenic arsenic nitrogen, arsenic oxygen, the difference in electro negativity increases, which means the bonds are becoming more polar. So I can eliminate choice B. Now let's move on to choice C where we have aluminum, phosphorus and sulfur and they are all bonded to chlorine. Now take a look on the periodic table. You can see that aluminum and chlorine are rather far apart. And as I move from aluminum to phosphorus now those atoms are getting a little closer together on the periodic table and I move from phosphorus to sulfur. So we can see that in choice C the difference in electro negativity between the bonded bonded atoms is actually decreasing. So as I go from Al to PCL to SC those bonds are becoming less polar and that is the correct answer arranged in order of decreasing polarity. Now just to be thorough let's go ahead and confirm that choice D is incorrect. So we have antimony or SB bonded to iodine and then bromine and then chlorine. But as I move from iodine to bromine to chlorine the electro negativity of the atom is increasing which means that the difference in electro negativity between the bonded atoms is increasing. So in choice D those atoms are becoming more polar not less polar. So the correct answer to question three is choice C as those three bonded atoms with their decreasing difference in electro negativity are becoming less polar. Now let's move on to the review for topic 2.2 intramolecular force and potential energy. Here are some essential knowledge statements from the AP chemistry course and exam description. A graph of potential energy versus the distance between the atoms is a useful representation for describing the interactions between atoms. Such graphs illustrate both the equilibrium bond length, the separation between atoms at which the potential energy is lowest and the bond energy, the energy required to separate the atoms. In a covealent bond, the bond length is influenced by both the size of the atom's core and the bond order that is single, double, triple. Bonds with a higher order are shorter and have larger bond energies. Kulum's law can be used to understand the strength of interactions between cations ions and annions because the interaction strength is proportional to the charge on each ion. Larger charges lead to stronger interactions because the interaction strength increases as the distance between the centers of the ions or nuclei decreases. Smaller ions lead to stronger interactions. Now let's move on to the practice for topic 2.2. This is question four. In question four, it says, "A potential energy diagram for Cl2 is shown in the following graph." And in part A, it says, based on the graph, identify the bond length and the bond energy for Cl2. I'd like to show you in general how to interpret these types of graphs. So you'll notice that the Yaxis is normally labeled as energy or potential energy in units of KJ per mole. And then on the x-axis we can either see bond length or interuclear distance the distance between the nuclei of two atoms and that's normally measured in pometers. So on one extreme you have two atoms in which the distance between the nuclei are very far apart and the potential energy is zero. So why is the potential energy zero if the atoms are far apart? It's because there is neither attraction or repulsion between the two atoms. So the potential energy is zero. Now let's go to the other extreme on the left side of this graph. If those two atoms are very close together, so the distance between the nuclei is very small. Now we get repulsion between the positively charged nuclei and that causes a level of unstable behavior and the potential energy goes way up. So where is the minimum in potential energy where the two atoms form a stable bond? Well, it's right here at that minimum in potential energy at the bottom of that curve. So what we can do if we see a curve like this is that along the x axis that would be the bond length of that particular bond. And then from zero to the bottom or the minimum in potential energy that is the bond energy and that's the energy that is either released when the bond is formed but that value can also be the energy that is absorbed when the bond is broken. Now let's go back to the diagram in question four. So looking along the x axis for that minimum in potential energy we could estimate that the bond length is approximately 200 pometers and then from zero down to the bottom of that curve it looks like it's somewhere between 200 and 300 kJ per mole. So estimating from the graph looking at the axis I would say that's around 250 kJ per mole. Now let's move on to part B. of question four. So in part B of question four, it says a student finds that the average aluminum chlorine bond length is 220 pomeres. I'm going to go ahead and draw a red line on the graph at 220 pometers. So pay close attention to the scale on the x axis. So 220 pometers would be right here. And then it says that the average bond energy for this bond is 425 kJ per mole. So somewhere between 400 and 500 I would say that 425 kJ per mole would be located right here on the y ais. Now what I'm going to do is put a green dot on the graph at the intersection of those two red lines. What we have to do is draw the potential energy curve for the aluminum chlorine bond on this graph. So notice that what I'm doing on the left side is I'm following that same pattern where the potential energy increases as the atoms get closer and closer together. So I'm just modeling the curve that we can already see drawn. Now going where the atoms are farther apart. You can see I'm trying to get the potential energy to match that curve. So around zero when the atoms are far apart. I'm now going to connect these green dots with a green line. And this is my answer to part B. I have sketched the potential energy curve for the average aluminum chlorine bond with a bond length of 220 pometers and a bond energy of 425 kJ per mole. In question five, we are given a potential energy versus bond length diagram for an HCl molecule. I'm going to zoom in a little bit more closely so we can identify the bond length and the bond energy for this HCl molecule. So it looks like the bond length would be somewhere between 100 and 200 pometers. I think it's less than 150. I'm going to estimate that the bond length is approximately 125 pometers. And now I'll take a look at the bond energy. I would say the bond energy is somewhere between 400 and 500. Looks like that would be 450. So I'm going to estimate that the bond energy is approximately 430 K per mole. So what I have is my estimated value for bond length and bond energy for the HCl molecule. Now the question asks which of the following graph shows the potential energy versus bond length for an HI molecule. Now chlorine versus iodine. You may remember for the periodic trend in atomic radius that atomic radius tends to increase as you move from top to bottom down a group. Let's compare chlorine and iodine. we would expect that the HC bond would be shorter and the HI bond would be longer. So since iodine based on periodic trends has a larger atomic radius than chlorine, the HI bond should have a longer bond length than the HCl bond. And as you might imagine, a longer bond will actually be easier to break. So, because the HI bond is longer and should be easier to break, the HI bond would have a lower bond energy. I'm now going to put on this graph my estimated curve where the bond length would be shifted over to the right and that the bond energy would be less. So, as you can see, that red curve is what I'm expecting for the HI molecule. Now, let's consider the four choices in question five. Now for choice A and C you can see that the location on the X axis is approximately 100 pometers but that's actually shorter than 125. I'm looking for a longer bond length and I see a shorter bond length in choices A and C. So I'm going to go ahead and eliminate choices A and C. I'm also looking for a lower bond energy. And in choice D, it looks like the bond energy is almost 500 kJ per mole. So a larger bond energy is not what I'm looking for. So I can eliminate choice D. So the correct answer to this question is choice B. And you can see I'm looking for a longer bond length and a lower bond energy for the HI bond. So that's why choice B is the correct answer. Now let's take a look at question six. In question six, it says the structural formulas of three different compounds are shown in the table below. So from left to right, we have C2H2 and as you can see there is a triple bond between the carbon atoms and then we have C2H4 and you can see a double bond between the carbon atoms and then finally C2H6 and there is a single bond between the carbon atoms. Part A says which compound contains the shortest carbontocarbon bond and part B says which compound contains the lowest value for its carbontocarbon bond energy. So this essential knowledge statement says that bonds with a higher order are shorter and have larger bond energies. So this CC triple bond in the compound C2H2 has a bond order of three. it is a shorter bond and has larger bond energy, which means it's stronger and requires more energy to break that bond. And then on the other side of the table, the CC single bond in the compound C2H6 has a bond order of one. That is a longer bond length and a smaller bond energy, which means it is a weaker bond and is easier to break. So now we can answer parts A and B of question six. Which compound contains the shortest carbontocarbon bond? Well, that would be C2H2 with that triple bond between the carbons. And then which compound contains the lowest value for its carbontocarbon bond energy? The weakest bond that's the easiest to break would be the single bond in C2H6. All right. Now, let's take a look at question seven. Question seven asks, which of the following compounds has the strongest colombic attractions between its ions in the solid state? So, let's take a look at both the charges of the ions as well as the relative sizes of the ions. Because the interaction strength is proportional to the charge on each ion, larger charges lead to stronger interactions. And because the interaction strength increases as the distance between the centers of the ions decreases, smaller ions lead to stronger interactions. Now looking at these four ionic compounds, we see calcium and barerium. Calcium and berium are located in group two. And the typical charge on a monatomic ion in group two is 2+. Then we have either oxygen or chlorine. And the typical charge on the monatomic ion oxygen would be -2 or 2 minus. And then chlorine would be -1 or 1 minus. So now + 2 -2 + 2 -1 + 2 - 2 + 2 -1. But larger ion charges result in stronger colombic attractions. So let's go ahead and eliminate choices B and D. And then smaller ion sizes result in stronger colombic attractions. Now calcium has fewer occupied electron shells. Barerium has more occupied electron shells. Calcium is smaller. Barerium is larger. So we're going to go with the smaller ionic size. And that would be calcium versus barerium. So the correct answer to question seven is choice A. Now let's talk about question eight which also involves an attraction between ions. In question eight it says the energy required to separate the ions in the sodium chloride crystal lice into individual sodium plus and chloride minus ions in the gas phase as represented in the table above. This is known as the lattice energy of sodium chloride. Which of the following correctly identifies a compound with a lattice energy that is greater than 786 kJ per mole and provides the correct justification. So we are looking for a compound with stronger colombic attractions between the ions and our choices are either potassium chloride or magnesium chloride. Now since each of the compounds listed contains the chloride ion, let's focus our attention on the cation or the positive ion. And again greater than 786 kJ per mole means that they're looking for a stronger colombic attraction between the ions to answer this question. So we have to ask ourselves which ion potassium. So K is +1 or magnesium Mg is +2 would result in stronger attractions. Larger charges lead to stronger interactions. So it's going to be not potassium which is which is + one but rather magnesium which is plus two. Now we know that the correct answer to question 8 is either C or D. But in choice C, the justification says the mass of the magnesium ion is greater. And in choice D, it says the charge on the magnesium ion. Well, it's definitely not about the mass. It is about the magnitude of the charge. So the correct answer to question A is choice D. The charge on the magnesium ion is greater than the charge on the sodium ion. And this results in stronger colombic attractions between the ions in the magnesium chloride crystal lattice. Now let's move on to the portion of the lesson where I give you advice to help you avoid common mistakes. My first piece of advice is what I've said before in my previous lessons. Read the question carefully and make sure you are answering the question that is being asked with respect to periodic trends in electro negativity. Electro negativity values tend to increase from left to right across a period and decrease going from top to bottom down a group. With respect to the polarity of a bond in a polar covealent bond, the atom with the higher electro negativity value attracts electrons more strongly toward itself and then carries a partial negative charge. And the relative difference in electro negativity between two atoms can be used to compare different bonds. The greater the difference in electro negativity, the more polar the bond or the greater the dipole moment and the greater the ionic character of the bond. A graph that shows energy on the y-axis and interuclear distance or bond length on the x-axis is used to illustrate the bond length x-axis and bond energy y-axis. Pay close attention to the axis labels so that you can read the correct value for both bond length in pometers and bond energy in KJ per mole. Now a bond energy value should be recorded as a positive number and the bond energy can be defined as either the energy required to break the bond or the energy released when the bond is formed. It is the absolute value of the minimum energy value of the curve. And when comparing two different single covealent bonds, smaller atoms result in shorter bonds, that is smaller distance between the nuclei. Larger atoms result in longer bonds, that is larger distance between the nuclei. The higher the bond order, the stronger and shorter the bonds tend to be. So assuming that different bonds containing the same atoms are compared for example carbonarbon single double and triple triple bonds or a bond order of three are stronger and shorter than double bonds with a bond order of two and double bonds bond order of two are stronger and shorter than single bonds bond order of one. Now, when comparing two different ionic compounds, a greater charge on an ion should result in stronger colombic attractive forces between the oppositely charged ions in the crystal lice. And then finally, a smaller ionic radius should result in a smaller distance between the ions and the crystal lice and stronger colombic attractive forces between the oppositeely charged ions. So now we have come to the end of the lesson. In my next lesson, I'll give you the chance to review and practice topics 2.3 and topic 2.4. If you enjoyed this lesson, I invite you to like this video and subscribe to my YouTube channel. Check out the links in the video description area, which include the link to my YouTube playlist organized by AP Chemistry CED units. You can add your comments and questions below this video and you can also email me with your comments and questions at a chemistry review and [email protected]. Thank you for letting me be your guide on the side as you strive toward that five on the AP chemistry exam. Keep studying and I'll see you next time on AP Chemistry Review and Practice. [Music] Heat.

I'm going to go ahead and draw a red line on the graph at 220 pometers. So pay close attention to the scale on the x-axis. So 220 pometers would be right here. And then it says that the average bond energy for this bond is 425 kJ per mole. So somewhere between 400 and 500 I would say that 425 kJ per mole would be located right here on the y ais. Now what I'm going to do is put a green dot on the graph at the intersection of those two red lines. [Music] Welcome to AP Chemistry Review and Practice. My name is Michael Farbaugh and I'll be your guide on the side as you strive toward that five on the AP Chemistry exam. Each episode will include answers and explanations to practice problems along with advice to help you avoid common mistakes. So, are you ready to review and practice for the AP Chemistry exam? I hope so, because this lesson starts right now. [Music] Welcome back to AP Chemistry Review and Practice. In this lesson, I'll give you the chance to review and practice topics 2.1 and 2.2. As you can see on this slide, these topics fall under unit 2 in the AP Chemistry curriculum, compound structure and properties. You can add your comments and questions below this video, and you can also email me with your comments and questions at a chemistry review and practice atgmail.com. Check out the links in the video description area, which include the link to the packet that accompanies this video. Now let's move on to the review for topic 2.1 types of chemical bonds. Here are the essential knowledge statements from the AP chemistry course and exam description. Electro negativity values for the representative elements increase going from left to right across a period and decrease going down a group. These trends can be understood qualitatively through the electronic structure of the atoms, the shell model and Kulum's law. Veence electrons shared between atoms of similar electro negativity constitute a non-polar covealent bond. For example, bonds between carbon and hydrogen are effectively non-polar even though carbon is slightly more electrogative than hydrogen. Veence electrons shared between atoms of unequal electro negativity constitute a polar covealent bond. The atom with a higher electro negativity will develop a partial negative charge relative to the other atom in the bond. In single bonds, greater differences in electro negativity lead to greater bond dipoles. All polar bonds have some ionic character and the difference between ionic and covealent bonding is not distinct but rather a continuum. The difference in electro negativity is not the only factor in determining if a bond should be designated as ionic or covealent. Generally bonds between a metal and non-metal are ionic and bonds between two non-metals are covealent. Examination of properties of a compound is the best way to characterize the type of bonding in a metallic solid. The veence electrons from the metal atoms are considered to be deoized and not associated with any individual atom. Now let's move on to the practice for topic 2.1. This is question one. It says a student assigned partial charges to each atom in each of the bonds shown below. Based on the electro negativity values of the atoms involved in each bond, circle all of the bonds for which the partial charges were labeled correctly by the student. So what we need to do to answer this question is consider that the atom with the higher electro negativity will develop a partial negative charge relative to the other atom in the bond. So these are the electro negativity values for various elements and you can see that the general trend is that electro negativity values tend to increase from left to right and they decrease from top to bottom. Now even though I'm showing you the actual electro negativity values on this slide, remember that you don't need to memorize these values for the AP chemistry exam. All you need to do is consider the general trends in electro negativity when comparing two different atoms. So the first two atoms that we're going to compare are arsenic as s and oxygen O. So we can see that oxygen is further to the top right corner of the periodic table compared to arsenic. So just based on the general trend, oxygen has a higher electro negativity than arsenic has. That means that the oxygen atom should have a partial negative charge. Now we'll see if the student assigned the partial charges correctly. Now they're putting a positive charge on oxygen instead of a negative charge. So that is incorrect. And now let's take a look at silicon and chlorine. Now we can see that silicon and chlorine are located in the same period. But since chlorine is further to the right, chlorine would have a higher electro negativity than silicon. So the chlorine atom should have a partial negative charge. Let's see if the student did this particular charge assignment correctly. And yes, there is a partial negative charge on the chlorine atom. Now let's move on to phosphorus and nitrogen. We can see that they are located in the same group. But since nitrogen is further to the top of the group, nitrogen should have a higher electro negativity than phosphorus has. So the nitrogen atom should have a partial negative charge and we can see that the student did also assign those charges correctly. Now moving on to a comparison of phosphorus and florine. So florine being further to the top right has a higher electro negativity than phosphorus has. So the florine atom should have a partial negative charge which the student also assigned correctly. Our next comparison is bromine and selenium. Now bromine is further to the right in that period compared to selenium. So bromine has a higher electro negativity than selenium. The bromine atom should have a partial negative charge which the student did not assign correctly. And then finally for chlorine and iodine which are located in the same group. Chlorine being further to the top of that group has a higher electro negativity than iodine. So the chlorine atom should have a partial negative charge which the student did not assign correctly. So these three circled bonds are the correct answers for question one in which all of the partial charges were labeled correctly. The atom with the higher electro negativity carries that partial negative charge in a polar covealent bond. Now let's take a look at question two. In question two, we are given three different chemical formulas and the question says which of the following correctly ranks the substances listed above in order of increasing ionic character in their bonding. So each compound contains the element chlorine CL, but we have silicon SI, magnesium Mg, and sulfur S. Now take a look on this slide. You can see that I've highlighted magnesium, silicon, sulfur, and chlorine which are all located in the same period of the periodic table. And you can also see that I've labeled these elements as either a metal, a non-metal or a metaloid. Now there is an important difference between a covealent bond and an ionic bond. And the usual method used in chemistry to classify a bond as either ionic or covealent is the following. metal and non-metal would lead to an ionic bond and two non-metals would lead to a covealent bond. Now this is a simple criterion that works well for many examples. However, the nature of a chemical bond is more complicated than this. In reality, all polar bonds have some ionic character and there is a continuum between the extremes of covealent bonding and ionic bonding. Now take a look on this slide and you'll see a link in the video description area to the FET simulation on molecule polarity. So what I'm doing is I'm moving that slider at the bottom and you can see that atom B which is labeled in green I'm increasing the electro negativity of that atom. So as I increase the electro negativity of atom B, I'm also increasing the magnitude of the bond dipole which is represented by that black arrow. So as the difference in electro negativity increases between the atoms, the bond character changes from more covealent to more ionic. So the ionic character of a bond increases with increasing differences between the electro negativities of the bonded elements. And that's how we're going to answer question two. So again all polar bonds have some ionic character and the difference between ionic and coalent bonding is not distinct but rather a continuum. Let's consider that the difference in electro negativity between magnesium and chlorine. You can see how far apart they are in the periodic table is greater than the difference in electro negativity between silicon and chlorine. Again, the difference in electro negativity between silicon and chlorine is greater than that between sulfur and chlorine. So, the sc2 compound has the lowest amount of ionic character and the mg2 compound has the greatest amount of ionic character. And it turns out that magnesium and chlorine are a metal and a non-metal. So magnesium and chlorine is more likely to form an ionic bond. So again, the correct answer is choice B. The ionic character of a bond increases with increasing differences between the electro negativities of the bonded elements. Now let's take a look at question three. Question three says, which of the following bonds are arranged in order of decreasing polarity? So this time we're looking for a smaller difference in electro negativity. In single bonds, greater differences in electro negativity will lead to greater bond dipoles. The greater the difference in electro negativity, the more polar the bond is. But in question three, we're looking for decreasing polarity. So we would like to see a decreasing difference in the electro negativities of the bonded atoms. Let's start with choice A. We have oxygen, nitrogen, and boron. And they are all bonded to the element florine. And you can see on the periodic table that oxygen and florine are closer together. And as I move from oxygen to nitrogen, now those two atoms are a little farther apart. And as I move from nitrogen to boron, those atoms are even farther apart, which means that would lead to a difference in electro negativity that's increasing. So in choice A, those bonds are actually becoming more polar, not less polar. So we can eliminate choice A. Now moving on to choice B, we have arsenic as bonded to either phosphorus, nitrogen or oxygen. So as I move from phosphorus to nitrogen to oxygen, I'm increasing the electro negativity of the atom that is bonded to arsenic. So again just like in choice A, as I move from arsenic phosphorus, arsenic arsenic nitrogen, arsenic oxygen, the difference in electro negativity increases, which means the bonds are becoming more polar. So I can eliminate choice B. Now let's move on to choice C where we have aluminum, phosphorus and sulfur and they are all bonded to chlorine. Now take a look on the periodic table. You can see that aluminum and chlorine are rather far apart. And as I move from aluminum to phosphorus now those atoms are getting a little closer together on the periodic table and I move from phosphorus to sulfur. So we can see that in choice C the difference in electro negativity between the bonded bonded atoms is actually decreasing. So as I go from Al to PCL to SC those bonds are becoming less polar and that is the correct answer arranged in order of decreasing polarity. Now just to be thorough let's go ahead and confirm that choice D is incorrect. So we have antimony or SB bonded to iodine and then bromine and then chlorine. But as I move from iodine to bromine to chlorine the electro negativity of the atom is increasing which means that the difference in electro negativity between the bonded atoms is increasing. So in choice D those atoms are becoming more polar not less polar. So the correct answer to question three is choice C as those three bonded atoms with their decreasing difference in electro negativity are becoming less polar. Now let's move on to the review for topic 2.2 intramolecular force and potential energy. Here are some essential knowledge statements from the AP chemistry course and exam description. A graph of potential energy versus the distance between the atoms is a useful representation for describing the interactions between atoms. Such graphs illustrate both the equilibrium bond length, the separation between atoms at which the potential energy is lowest and the bond energy, the energy required to separate the atoms. In a covealent bond, the bond length is influenced by both the size of the atom's core and the bond order that is single, double, triple. Bonds with a higher order are shorter and have larger bond energies. Kulum's law can be used to understand the strength of interactions between cations ions and annions because the interaction strength is proportional to the charge on each ion. Larger charges lead to stronger interactions because the interaction strength increases as the distance between the centers of the ions or nuclei decreases. Smaller ions lead to stronger interactions. Now let's move on to the practice for topic 2.2. This is question four. In question four, it says, "A potential energy diagram for Cl2 is shown in the following graph." And in part A, it says, based on the graph, identify the bond length and the bond energy for Cl2. I'd like to show you in general how to interpret these types of graphs. So you'll notice that the Yaxis is normally labeled as energy or potential energy in units of KJ per mole. And then on the x-axis we can either see bond length or interuclear distance the distance between the nuclei of two atoms and that's normally measured in pometers. So on one extreme you have two atoms in which the distance between the nuclei are very far apart and the potential energy is zero. So why is the potential energy zero if the atoms are far apart? It's because there is neither attraction or repulsion between the two atoms. So the potential energy is zero. Now let's go to the other extreme on the left side of this graph. If those two atoms are very close together, so the distance between the nuclei is very small. Now we get repulsion between the positively charged nuclei and that causes a level of unstable behavior and the potential energy goes way up. So where is the minimum in potential energy where the two atoms form a stable bond? Well, it's right here at that minimum in potential energy at the bottom of that curve. So what we can do if we see a curve like this is that along the x axis that would be the bond length of that particular bond. And then from zero to the bottom or the minimum in potential energy that is the bond energy and that's the energy that is either released when the bond is formed but that value can also be the energy that is absorbed when the bond is broken. Now let's go back to the diagram in question four. So looking along the x axis for that minimum in potential energy we could estimate that the bond length is approximately 200 pometers and then from zero down to the bottom of that curve it looks like it's somewhere between 200 and 300 kJ per mole. So estimating from the graph looking at the axis I would say that's around 250 kJ per mole. Now let's move on to part B. of question four. So in part B of question four, it says a student finds that the average aluminum chlorine bond length is 220 pomeres. I'm going to go ahead and draw a red line on the graph at 220 pometers. So pay close attention to the scale on the x axis. So 220 pometers would be right here. And then it says that the average bond energy for this bond is 425 kJ per mole. So somewhere between 400 and 500 I would say that 425 kJ per mole would be located right here on the y ais. Now what I'm going to do is put a green dot on the graph at the intersection of those two red lines. What we have to do is draw the potential energy curve for the aluminum chlorine bond on this graph. So notice that what I'm doing on the left side is I'm following that same pattern where the potential energy increases as the atoms get closer and closer together. So I'm just modeling the curve that we can already see drawn. Now going where the atoms are farther apart. You can see I'm trying to get the potential energy to match that curve. So around zero when the atoms are far apart. I'm now going to connect these green dots with a green line. And this is my answer to part B. I have sketched the potential energy curve for the average aluminum chlorine bond with a bond length of 220 pometers and a bond energy of 425 kJ per mole. In question five, we are given a potential energy versus bond length diagram for an HCl molecule. I'm going to zoom in a little bit more closely so we can identify the bond length and the bond energy for this HCl molecule. So it looks like the bond length would be somewhere between 100 and 200 pometers. I think it's less than 150. I'm going to estimate that the bond length is approximately 125 pometers. And now I'll take a look at the bond energy. I would say the bond energy is somewhere between 400 and 500. Looks like that would be 450. So I'm going to estimate that the bond energy is approximately 430 K per mole. So what I have is my estimated value for bond length and bond energy for the HCl molecule. Now the question asks which of the following graph shows the potential energy versus bond length for an HI molecule. Now chlorine versus iodine. You may remember for the periodic trend in atomic radius that atomic radius tends to increase as you move from top to bottom down a group. Let's compare chlorine and iodine. we would expect that the HC bond would be shorter and the HI bond would be longer. So since iodine based on periodic trends has a larger atomic radius than chlorine, the HI bond should have a longer bond length than the HCl bond. And as you might imagine, a longer bond will actually be easier to break. So, because the HI bond is longer and should be easier to break, the HI bond would have a lower bond energy. I'm now going to put on this graph my estimated curve where the bond length would be shifted over to the right and that the bond energy would be less. So, as you can see, that red curve is what I'm expecting for the HI molecule. Now, let's consider the four choices in question five. Now for choice A and C you can see that the location on the X axis is approximately 100 pometers but that's actually shorter than 125. I'm looking for a longer bond length and I see a shorter bond length in choices A and C. So I'm going to go ahead and eliminate choices A and C. I'm also looking for a lower bond energy. And in choice D, it looks like the bond energy is almost 500 kJ per mole. So a larger bond energy is not what I'm looking for. So I can eliminate choice D. So the correct answer to this question is choice B. And you can see I'm looking for a longer bond length and a lower bond energy for the HI bond. So that's why choice B is the correct answer. Now let's take a look at question six. In question six, it says the structural formulas of three different compounds are shown in the table below. So from left to right, we have C2H2 and as you can see there is a triple bond between the carbon atoms and then we have C2H4 and you can see a double bond between the carbon atoms and then finally C2H6 and there is a single bond between the carbon atoms. Part A says which compound contains the shortest carbontocarbon bond and part B says which compound contains the lowest value for its carbontocarbon bond energy. So this essential knowledge statement says that bonds with a higher order are shorter and have larger bond energies. So this CC triple bond in the compound C2H2 has a bond order of three. it is a shorter bond and has larger bond energy, which means it's stronger and requires more energy to break that bond. And then on the other side of the table, the CC single bond in the compound C2H6 has a bond order of one. That is a longer bond length and a smaller bond energy, which means it is a weaker bond and is easier to break. So now we can answer parts A and B of question six. Which compound contains the shortest carbontocarbon bond? Well, that would be C2H2 with that triple bond between the carbons. And then which compound contains the lowest value for its carbontocarbon bond energy? The weakest bond that's the easiest to break would be the single bond in C2H6. All right. Now, let's take a look at question seven. Question seven asks, which of the following compounds has the strongest colombic attractions between its ions in the solid state? So, let's take a look at both the charges of the ions as well as the relative sizes of the ions. Because the interaction strength is proportional to the charge on each ion, larger charges lead to stronger interactions. And because the interaction strength increases as the distance between the centers of the ions decreases, smaller ions lead to stronger interactions. Now looking at these four ionic compounds, we see calcium and barerium. Calcium and berium are located in group two. And the typical charge on a monatomic ion in group two is 2+. Then we have either oxygen or chlorine. And the typical charge on the monatomic ion oxygen would be -2 or 2 minus. And then chlorine would be -1 or 1 minus. So now + 2 -2 + 2 -1 + 2 - 2 + 2 -1. But larger ion charges result in stronger colombic attractions. So let's go ahead and eliminate choices B and D. And then smaller ion sizes result in stronger colombic attractions. Now calcium has fewer occupied electron shells. Barerium has more occupied electron shells. Calcium is smaller. Barerium is larger. So we're going to go with the smaller ionic size. And that would be calcium versus barerium. So the correct answer to question seven is choice A. Now let's talk about question eight which also involves an attraction between ions. In question eight it says the energy required to separate the ions in the sodium chloride crystal lice into individual sodium plus and chloride minus ions in the gas phase as represented in the table above. This is known as the lattice energy of sodium chloride. Which of the following correctly identifies a compound with a lattice energy that is greater than 786 kJ per mole and provides the correct justification. So we are looking for a compound with stronger colombic attractions between the ions and our choices are either potassium chloride or magnesium chloride. Now since each of the compounds listed contains the chloride ion, let's focus our attention on the cation or the positive ion. And again greater than 786 kJ per mole means that they're looking for a stronger colombic attraction between the ions to answer this question. So we have to ask ourselves which ion potassium. So K is +1 or magnesium Mg is +2 would result in stronger attractions. Larger charges lead to stronger interactions. So it's going to be not potassium which is which is + one but rather magnesium which is plus two. Now we know that the correct answer to question 8 is either C or D. But in choice C, the justification says the mass of the magnesium ion is greater. And in choice D, it says the charge on the magnesium ion. Well, it's definitely not about the mass. It is about the magnitude of the charge. So the correct answer to question A is choice D. The charge on the magnesium ion is greater than the charge on the sodium ion. And this results in stronger colombic attractions between the ions in the magnesium chloride crystal lattice. Now let's move on to the portion of the lesson where I give you advice to help you avoid common mistakes. My first piece of advice is what I've said before in my previous lessons. Read the question carefully and make sure you are answering the question that is being asked with respect to periodic trends in electro negativity. Electro negativity values tend to increase from left to right across a period and decrease going from top to bottom down a group. With respect to the polarity of a bond in a polar covealent bond, the atom with the higher electro negativity value attracts electrons more strongly toward itself and then carries a partial negative charge. And the relative difference in electro negativity between two atoms can be used to compare different bonds. The greater the difference in electro negativity, the more polar the bond or the greater the dipole moment and the greater the ionic character of the bond. A graph that shows energy on the y-axis and interuclear distance or bond length on the x-axis is used to illustrate the bond length x-axis and bond energy y-axis. Pay close attention to the axis labels so that you can read the correct value for both bond length in pometers and bond energy in KJ per mole. Now a bond energy value should be recorded as a positive number and the bond energy can be defined as either the energy required to break the bond or the energy released when the bond is formed. It is the absolute value of the minimum energy value of the curve. And when comparing two different single covealent bonds, smaller atoms result in shorter bonds, that is smaller distance between the nuclei. Larger atoms result in longer bonds, that is larger distance between the nuclei. The higher the bond order, the stronger and shorter the bonds tend to be. So assuming that different bonds containing the same atoms are compared for example carbonarbon single double and triple triple bonds or a bond order of three are stronger and shorter than double bonds with a bond order of two and double bonds bond order of two are stronger and shorter than single bonds bond order of one. Now, when comparing two different ionic compounds, a greater charge on an ion should result in stronger colombic attractive forces between the oppositely charged ions in the crystal lice. And then finally, a smaller ionic radius should result in a smaller distance between the ions and the crystal lice and stronger colombic attractive forces between the oppositeely charged ions. So now we have come to the end of the lesson. In my next lesson, I'll give you the chance to review and practice topics 2.3 and topic 2.4. If you enjoyed this lesson, I invite you to like this video and subscribe to my YouTube channel. Check out the links in the video description area, which include the link to my YouTube playlist organized by AP Chemistry CED units. You can add your comments and questions below this video and you can also email me with your comments and questions at a chemistry review and practice@gmail.com. Thank you for letting me be your guide on the side as you strive toward that five on the AP chemistry exam. Keep studying and I'll see you next time on AP Chemistry Review and Practice. [Music] Heat.

Transcript for:AP Chemistry Bonding Overview

Transcript for:
AP Chemistry Bonding Overview